Question

Find the coordinates of the point of trisection of the line segment joining the points A=(5,-3) and B(2,-9)

Answer 1

let the points of trisection be P(X,y) and R(a,b)

coordinates of P(7/2, -6)

coordinates of R (3, -15/2)

by using formula (x1 +x2) /2 and (y1+y2)/2

u will get the answer....

HOPE YOU UNDERSTAND ☺️

Answer 2

Concept: In coordinate geometry, the Section formula is used to determine the internal or external ratio at which a line segment is divided by a point. It is used to determine a triangle's centroid, incenter, and excenter. It is employed in physics to identify equilibrium locations, system centres of mass, etc.

Given: the points A(5,-3) and B(2,-9)

To find: the coordinates of the point of trisection of the line segment joining the points A(5,-3) and B(2,-9)

Solution:

The given points are A(5, -3) and B(2, -9).

Let P(x, y) and Q(x', y') be the points of trisection of the line segment joining the points A(5,-3) and B(2,-9).

First we find the coordinates of P using the section formula,

$(\frac{mx_{2} + nx_1 }{m + n} , \frac{my_{2} + ny_1 }{m + n})$

Here, m and n are the ratios in which the line segment joining the points A(5,-3) and B(2,-9) are divided.

Point A(5, -3) is taken as $x_{1} y_{1}$ and point B(2, -9) is taken as $x_{2} y_{2}$.

Therefore, for co-ordinate P(x, y) the ration m:n is 1:2

Hence,

$P(x, y) = (\frac{1* 2 + 2* 5}{1+2} , \frac{1*(-9) + 2*(-3) }{1+2})\\\\P(x, y) = (\frac{2 + 10}{3} , \frac{(-9) + (-6) }{3})\\\\P(x, y) = (\frac{12}{3} , \frac{-15 }{3})\\\\P(x, y) = (4, -5)$

Now, for finding the co-ordinates of Q(x', y'), the ration m:n becomes 2:1

Hence,

$Q(x', y') = (\frac{2* 2 + 1* 5}{2+1} , \frac{2*(-9) + 1*(-3) }{2+1})\\\\Q(x', y') = (\frac{4 + 5}{3} , \frac{(-18) + (-3) }{3})\\\\Q(x', y') = (\frac{9}{3} , \frac{-21 }{3})\\\\Q(x', y') = (3, -7)$

Hence, the coordinates of the point of trisection of the line segment joining the points A(5,-3) and B(2,-9) are P(4, -5) and Q(3, -7).

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